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Math Help - Equliateral Triangle inscribed in circle

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    Equliateral Triangle inscribed in circle

    Equilateral triangle ABC is inscribed in a circle. The angle bisector at A intersects the circle again at P (AP is the diameter). If the length of segment AC is 10, find AP - (BP + CP).


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  2. #2
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    Quote Originally Posted by hachiko View Post
    Equilateral triangle ABC is inscribed in a circle. The angle bisector at A intersects the circle again at P (AP is the diameter). If the length of segment AC is 10, find AP - (BP + CP).
    Let O be the centre of the circle. Can you see why OP = BP = CP? What does that tell you about AP (BP + CP)?
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  3. #3
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    Hello, hachiko!

    Another approach . . .


    Equilateral triangle ABC is inscribed in a circle.
    The angle bisector at A intersects the circle again at P ( AP is adiameter).
    If the length of segment AC is 10, find: .  AP - (BP + CP)
    Code:
                    A
                  * o *
              *    /|\    *
            *     / | \     *
           *     /  |  \ 10  *
                /   |30\
          *    /    |    \    *
          *   /     |     \   *
          *  /      |      \  *
            /       |       \
         B o- - - - + - - - -o C
            *       |     * *
              * 60 |  *  *
                  * o *
                    P

    We have: inscribed equilateral triangle ABC\!:\;\;AC = 10
    . . with diameter AP,\;\;\angle CAP = 30^o.
    Draw chord CP.

    \Delta ACP is inscribed in a semicircle.
    . . Hence: . \angle ACP = 90^o,\;\;\angle APC = 60^o


    We have: . CP = x,\;\;AC = 10,\;\;AP = 2x .(30-60 right triangle)

    Pythagorus says: . CP^2 + AC^2 \:=\:AP^2 \quad\Rightarrow\quad x^2 + 10^2 \:=\:(2x)^2

    . . x^2 + 100 \:=\:4x^2 \quad\Rightarrow\quad 3x^2 \:=\:100 \quad\Rightarrow\quad x^2 \:=\:\frac{100}{3}\quad\Rightarrow\quad x \:=\:\frac{10\sqrt{3}}{3}

    Hence: . BP \,=\,CP \,=\,\frac{10\sqrt{3}}{3},\;\;AP \,=\,\frac{20\sqrt{3}}{3}


    Therefore: . AP - (BP + CP) \;\;=\;\;\frac{20\sqrt{3}}{3} - \left(\frac{10\sqrt{3}}{3} + \frac{10\sqrt{3}}{3}\right) \;\;=\;\;0

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